The equation of the straight line passing through the point $(4,3)$ and making intercepts on the co-ordinate…
The equation of the straight line passing through the point $(4,3)$ and making intercepts on the co-ordinate axes whose sum is $-1$ is
$\frac{x}{2}+\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$
$\frac{x}{2}-\frac{y}{3}=-1$ and $\frac{x}{-2}+\frac{y}{1}=-1$
$\frac{x}{2}+\frac{y}{3}=1$ and $\frac{x}{2}+\frac{y}{1}=1$
$\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$
Solution
$\frac{x}{a}+\frac{y}{b}=1$ where $a+b=-1$ and $\frac{4}{a}+\frac{3}{b}=1$
$\Rightarrow a=2, b=-3$ or $a=-2, b=1$
Hence $\frac{x}{2}-\frac{y}{3}=1$ and $\frac{x}{-2}+\frac{y}{1}=1$